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I start with an equilateral triangle with side perimeter three meters. I can define a Koch snowflake by the following sequence of figures. Starting with that triangle, produce the next figure by replacing the middle third of each line segment with the other two sides of an equilateral triangle whose third vertex is exterior to the previous figure. We know that the Hausdorff dimension of the limit is (log 4)/(log 3).

Can we say that the measure of the snowflake's perimeter is three of a unit that we may express as one meter to the power ((log 4)/(log 3))? Something like that? Why, or why not?

I thought of this while fishing for a weekly brain teaser puzzle to pose to my office mates. I think it's too hard!

minopret
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  • Compute the perimeter $P_n$ at stage $n$; you'll see that $(P_n)$ is an increasing unbounded sequence. Conversely, if $A_n$ is the area at stage $n$, the sequence $(A_n)$ is increasing and bounded. – egreg Oct 28 '13 at 16:52
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    Relevant I think? So, it is quite difficult indeed? http://dx.doi.org/10.1016/j.amc.2007.01.046 – minopret Oct 28 '13 at 18:04
  • It has a positive and finite Hausdorff measure, am I right? I haven't carefully read and thought through what Hausdorff measure is all about. – minopret Nov 09 '13 at 20:22
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    Assuming you accept the theorem that Hausdorff dimension coincides with self-similarity dimension when the latter is define, the $((\log4)/(\log3)+\varepsilon)$-dimensional Hausdorff measures are $0$ for all positive $\varepsilon$, and $\infty$ for all negative $\varepsilon$. But a priori we know nothing about the measure of the Koch snowflake itself. For example, a line is 1-dimensional, but has 1-d measure $\infty$ (meters), and http://mathoverflow.net/questions/35986/measure-0-sets-on-the-line-with-hausdorff-dimension-1 alludes to 1-dimensional sets with 1-d measure 0. – Mark S. Nov 09 '13 at 20:45

1 Answers1

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The 2007 paper you linked in a comment, Bounds of the Hausdorff measure of the Koch curve, contains information which is the most recent I could find. In it, and the related Bounds of Hausdorff measure of the Sierpinski gasket, both by Baoguo Jia, an approach to estimating Hausdorff measures of self-similar sets satisfying the open set condition (citing some books by Falconer on fractal geometry for some facts about the Hausdorff measure for the proof that this method works) is outlined. However, the method is not very effective (it is not easy to calculate good approximations with certainty).

In "Bounds of the Hausdorff measure of the Koch curve", the author proves that the $s$-dimensional (where $s=\log4/\log3$) Hausdorff measure of the Koch curve (base length 1) is bounded below by $$\left(2 \left(\frac{2 \sqrt{3}}{9}\right)^s\right) \exp\left(-\frac{12 s\sqrt{3}}{9}\right)=2\text{^}\left(-2-\frac{8}{\sqrt{3} \log (3)}+s\right)\approx0.0325239$$and bounded above by $$2 \left(\frac{2 \sqrt{3}}{9}\right)^s=2^{s-2}\approx0.599512\text.$$

At the end of the paper, they conjecture tighter bounds. Assuming their $6/81$ was meant to be ${\sqrt{876}}/{81}$ (the context makes this a reasonable typo), they conjecture a lower bound of $$\left(\frac{1}{122} 4^4 \left(\frac{\sqrt{876}}{81}\right)^s\right) \exp\left(-\frac{12 s\sqrt{3} }{3^5}\right)=\frac{1}{61}73^{s/2}*2\text{^}\left(s-\frac{8}{27 \sqrt{3} \log (3)}\right)\approx0.528786$$ and an upper bound of $$\frac{1}{122} 4^4 \left(\frac{\sqrt{876}}{81}\right)^s=\frac{1}{61}73^{s/2}*2^s\approx0.589052\text.$$

In short, if the horizontal base for the curve is $1\mathrm{m}$, and this paper is correct, then the $s$-dimensional Hausdorff measure (which is the appropriate way to measure the size of something a self-similar fractal like this) is definitely less than $0.6\mathrm{m}^{s}$, and probably more than $0.5\mathrm{m}^{s}$.

Mark S.
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